Question 890075: 1/3log_cx+3log_cy-4log_cx
how would you express this as a single logarithm and simplify if possible?
Thanks.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! if i understand this correctly, your expression is:
1/3 * logc(x) + 3*logc(y) - 4*logc(x)
this becomes:
logc(x^(1/3)) + logc(y^3) - logc(x^4)
this becomes:
logc((x^(1/3) * y^3) / x^4)
this becomes:
logc(y^3/x^(11/3))
the base doesn't really matter in this problem.
it could be anything.
for example, if you let c = 10, than you can use the LOG function of your calculator to confirm this simplification is correct.
simply choose an arbitrary value for x and y and solve using your calculator for the original equation and the final equation.
for example:
let x = 20 and y = 35 and c = 10
1/3 * logc(x) + 3*logc(y) - 4*logc(x) becomes:
1/3 * log10(20) + 3 * log10(35 - 4 * log10(20) which is equal to -.1382391844...
logc(y^3/x^(11/3)) becomes:
log10(35^3 / 20^(11/3)) = -.1382391844...
you get the same value from the original equation and from the final equation so the simplification is correct.
the logarithmic rules that were used are:
a*log(b) = log(b^a)
log(a) + log(b) = log(a*b)
log(a) / log(b) = log(a/b)
the division rules that were followed are:
x^a / x^b = x^(a-b)
x^-c = 1/x^c
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